Limit of $\sum_{r=1}^{n} \frac{r}{2^r}$ as $n\to \infty$ 
Consider the sequence $u_{n}=\sum\limits_{r=1}^{n} \frac{r}{2^r}$ with $n \ge 1$.
  Then the limit of $u_{n}$ as $ n \rightarrow \infty$ is ?

I actually treated each term as an AGP and got $$u_{n}=2-\frac{1}{2^{r+1}}-\frac{n}{2^{r+1}}$$ 
But how to get the limit?
 A: This is a well-known result due to Nicole Oresme
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n}=2
\end{align}
Here is the proof 
Consider the function
\begin{align*}
f(z)=\frac{1}{1-z}
\end{align*}
The power series expansion of $f(z)$ is
\begin{align}
f(z)=\sum_{n=1}^{\infty}z^n
\end{align}
Now take the derivative of $f$, which is
\begin{align*}
f'(z)=\frac{1}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n-1}
\end{align*}
Multiplying by $z$
\begin{align}
zf'(z)=\frac{z}{(1-z)^2}=\sum_{n=1}^{\infty}nz^{n}\tag{1}
\end{align}
Substituting $z=1/2$ in (1)
\begin{align}
\sum_{n=1}^\infty \frac{n}{2^n}=\frac{\frac{1}{2}}{\biggl(1-\frac{1}{2}\biggl)^2}=2
\end{align} 
A: Let me do this (a bit non-rigorously) without derivatives.
Let $S=\frac12+\frac24+\frac38+\dots+\frac{r}{2^r}+\cdots$
Then
$$\begin{align}
S&=\frac12+\frac24+\frac38+\frac4{16}+\cdots
\\&=\frac12+\frac12\left(\frac22+\frac34+\frac48+\cdots\right)
\\&=\frac12+\frac12\left(\left(\frac12+\frac12\right)+\left(\frac24+\frac14\right)+\left(\frac38+\frac18\right)+\cdots\right)
\\&=\frac12+\frac12\left(\left(\frac12+\frac24+\frac38+\cdots\right)+\left(\frac12+\frac14+\frac18+\cdots\right)\right)
\\&=\frac12+\frac12(S+1)
\end{align}$$
Solving gives $S=2$.
A: Hint:
$$\sum_{r=1}^\infty r a^r = a \frac{d}{da}\sum_{r=1}^\infty a^r$$
